Problem: $ABCD$ is a regular tetrahedron (right triangular pyramid).  If $M$ is the midpoint of $\overline{CD}$, then what is $\cos \angle AMB$?
Solution: The tetrahedron is shown below.  In order to find $\cos \angle AMB$, we build a right triangle with $\angle AMB$ among its angles.  The foot of the altitude from $A$ to face $BCD$ is the centroid, $G$, of triangle $BCD$.

[asy]
import three;
currentprojection = orthographic(1.5,1.1,-1);
triple A = (1,1,1);
triple B = (1,0,0);
triple C = (0,1,0);
triple D = (0,0,1);
draw(A--B--C--A);
draw(A--D,dashed);
draw(C--D--B,dashed);
label("$A$",A,NW);
label("$B$",B,W);
label("$C$",C,S);
label("$D$",D,NW);
triple M = (0,0.5,0.5);
draw(A--M--B,dashed);
label("$M$",M,NE);
triple G = B/3 + 2*M/3;
draw(A--G,dashed);
label("$G$",G,S);

[/asy]

Since $\overline{BM}$ is a median of $\triangle BCD$, point $G$ is on $\overline{BM}$ such that $GM = \frac13BM$.  Furthermore, we have $AM = BM$, so  \[\cos \angle AMB= \cos \angle AMG = \frac{GM}{AM} = \frac{(BM/3)}{BM}=\boxed{\frac{1}{3}}.\]